# Inverse trignometry; sec and tan

• Apr 6th 2009, 12:25 PM
tiar
Inverse trignometry; sec and tan
I'm so stuck on this. I have no clue where to start.

sec (tan^-1 4)

(the tan is an inverse with the -1, 4 is the number)
• Apr 6th 2009, 12:34 PM
stapel
Quote:

Originally Posted by tiar
I'm so stuck on this. I have no clue where to start.

Does this mean that you're not familiar with the meanings of the inverse trigonometric functions?

In case you're exaggerating a bit...

Quote:

Originally Posted by tiar
sec (tan[/SIZE][SIZE=4]^-1 4)

By definition, arctan(4) = A means tan(A) = 4 = 4/1.

Draw a right triangle. Label the non-right base angle as A. Label the "opposite" and "adjacent" sides as 4 and 1, respectively.

Use the Pythagorean Theorem to find the length of the hypotenuse.

Read the value of the sine from the triangle. (Wink)
• Apr 6th 2009, 12:39 PM
tiar
No, I'm not very familiar with them at all.

Okay, so the hyp. of the right triangle would be $\displaystyle \sqrt{17}$ thus sine would be $\displaystyle \frac{4}{\sqrt{17}}$?

And I need to know the quadrant it is in to know if it is negative or positive, correct?
I'm confused on how to do that.
• Apr 6th 2009, 12:40 PM
icemanfan
Quote:

Originally Posted by tiar
I'm so stuck on this. I have no clue where to start.

sec (tan^-1 4)

(the tan is an inverse with the -1, 4 is the number)

You are looking for the secant of the angle whose tangent is 4. Remember that in a right triangle, $\displaystyle x^2 + y^2 = r^2$. Imagine a right triangle in which one leg (y) is four times as long as the other (x). This is indicated by the tangent (y/x) = 4. Hence,

$\displaystyle \frac{y}{x} = 4$
$\displaystyle y = 4x$
$\displaystyle x^2 + (4x)^2 = r^2$
$\displaystyle x^2 + 16x^2 = r^2$
$\displaystyle 17x^2 = r^2$
$\displaystyle 17 = \frac{r^2}{x^2}$
$\displaystyle \sqrt{17} = \frac{r}{x}$

Since the secant is r/x, it is equal to $\displaystyle \sqrt{17}$.
• Apr 6th 2009, 12:44 PM
icemanfan
Quote:

Originally Posted by tiar
No, I'm not very familiar with them at all.

Okay, so the hyp. of the right triangle would be $\displaystyle \sqrt{17}$ thus sine would be $\displaystyle \frac{4}{\sqrt{17}}$?

And I need to know the quadrant it is in to know if it is negative or positive, correct?
I'm confused on how to do that.

If you go strictly by the definition of the arctangent, it's range is only defined for the first and fourth quadrants. Since the value is positive (4), the angle arctan(4) must be located in the first quadrant.